Earlier I asked this question on the Math Exchange but I'm looking for a physics point of view. How do you interpret an equation like $$x^n \delta(x) = 0, \qquad n\in \mathbb{N},$$ around $x=0$? Why does it suffice to show the integral of this expression is zero around the singularity to show the equality is valid? There are many definitions to the Dirac delta "function" so I get extremely confused with this notion. In other words, do you treat equality of a Dirac delta function and some other expression as an "integral equality" because just inputting the value doesn't make sense?

  • 1
    $\begingroup$ Why wouldn’t inputtinh the value make sense? $$0^n\del(0) = 0*1=0$$? $\endgroup$
    – nemo
    Oct 30, 2020 at 16:55
  • $\begingroup$ @IronicalCoffee $\delta(0)\neq 1$, it is formally infinite. $\endgroup$
    – J. Murray
    Oct 30, 2020 at 16:58
  • $\begingroup$ Oh right I was confused with the kronecker delta, well then I don't think this equation holds as zero times infinity is an undefined number $\endgroup$
    – nemo
    Oct 30, 2020 at 17:15
  • 1
    $\begingroup$ While the below answer is nice and removes your doubts somehow, it isn't physically motivated like you requested and would be a perfectly fine response on math stackexchange too. A physically motivated answer would use the idea that delta functions represent idealized notions like point particles and sudden/instantaneous impulses to systems. Therefore I don't think you've gotten your actual answer. $\endgroup$
    – Triatticus
    Oct 30, 2020 at 22:02

1 Answer 1


There may be many different definitions of the Dirac delta, but they all share the common feature that they only make sense inside an integral, and that

$$\int_{-\infty}^\infty f(x)\delta(x-a)dx = f(a)\qquad (*)$$

My preferred perspective is that the collection of symbols on the left-hand side of $(*)$ is defined to mean $f(a)$, in the sense that the integrand is not a true function so the integral is not meant to be taken literally.

Another point of view is that the Dirac delta is the limit of a sequence of increasingly tall and narrow Gaussians,

$$\delta(x) = \lim_{\epsilon\rightarrow 0^+} \frac{e^{-x^2/\epsilon}}{\sqrt{\epsilon\pi}} \qquad (**)$$

enter image description here

For $x\neq 0$, the above limit evaluates to zero. For $x=0$, the limit does not exist (or is $\infty$, if you prefer). However,

$$\lim_{\epsilon\rightarrow 0^+}\int \frac{e^{-x^2/\epsilon}}{\sqrt{\epsilon\pi}} dx = \lim_{\epsilon\rightarrow 0^+}1 = 1$$

With a bit of work, one can show that $$\lim_{\epsilon\rightarrow 0^+} \int f(x) \frac{e^{-(x-a)^2/\epsilon}}{\sqrt{\epsilon\pi}} dx = f(a)$$

Therefore, we are to understand the expression $(**)$ as a limit which is to be taken only after an integral sign has been applied.

In both cases, the Dirac delta is an object which is only well-defined inside an integral. As a result, $x^n\delta(x)$ is defined in the same way.

  • $\begingroup$ At last, my confusion is settled. Thank you! $\endgroup$
    – Darkenin
    Oct 30, 2020 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.